6.001: Structure and Interpretation of Computer Programs Symbols Quotation Relevant details of the reader Example of using symbols Alists Differentiation

Data Types in Lisp/Scheme Conventional Numbers (integer, real, rational, complex) –Interesting property in “real” Scheme: exactness Booleans: #t, #f Characters and strings: #\a, “Hello World!” Vectors: #(0 “hi” 3.7) Lisp-specific Procedures: value of +, result of evaluating ( (x) x) Pairs and Lists: (3. 7), ( ) Symbols: pi, +, MyGreatGrandMotherSue

Symbols So far, we’ve seen them as the names of variables But, in Lisp, all data types are first class Therefore, we should be able to –Pass symbols as arguments to procedures –Return them as values of procedures –Associate them as values of variables –Store them in data structures –E.g., (apple orange banana) appleorange banana

How do we refer to Symbols? Substitution Model’s rule of evaluation: Value of a symbol is the value it is associated with in the environment We associate symbols with values using the special form define –(define pi ) … but that doesn’t help us get at the symbol itself

Referring to Symbols Say your favorite color Say “your favorite color” In the first case, we want the meaning associated with the expression, e.g., red In the second, we want the expression itself, e.g., your favorite color We use quotation to distinguish our intended meaning

New Special Form: quote Need a way of telling interpreter: “I want the following object as whatever it is, not as an expression to be evaluated” (quote alpha) ;Value: alpha (define pi ) ;Value: "pi --> " pi ;Value: (quote pi) ;Value: pi (+ pi pi) ;Value: (+ pi (quote pi)) ;The object pi, passed as the first argument to integer->flonum, is not the correct type. (define fav (quote pi)) fav ;Value: pi

A data abstraction consists of: constructors selectors operations contract Review: data abstraction (define make-point (lambda (x y) (list x y))) (define x-coor (lambda (pt) (car pt))) (define on-y-axis? (lambda (pt) (= (x-coor pt) 0))) (x-coor (make-point )) =

Symbol: a primitive type constructors: None since really a primitive, not an object with parts Only way to “make one” is to type it –(or via string->symbol from character strings, but shhhh…) selectors None –(except symbol->string) operations: symbol?; type: anytype -> boolean (symbol? (quote alpha)) ==> #t eq? ; discuss in a minute R5RS shows the full riches of Scheme

What’s the difference between symbols and strings? Symbol Evaluates to the value associated with it by define Every time you type a particular symbol, you get the exact same one! Guaranteed. –Called interning E.g., (list (quote pi) (quote pi)) String Evaluates to itself Every time you type a particular string, it’s up to the implementation whether you get the same one or different ones. E.g., (list ”pi” ”pi”) or pi ”pi”

The operation eq? tests for the same object a primitive procedure returns #t if its two arguments are the same object very fast (eq? (quote eps) (quote eps)) ==> #t (eq? (quote delta) (quote eps)) ==> #f For those who are interested: ; eq?: EQtype, EQtype ==> boolean ; EQtype = any type except number or string One should therefore use = for equality of numbers, not eq?

Making list structure with symbols ((red 700) (orange 600) (yellow 575) (green 550) (cyan 510) (blue 470) (violet 400)) (list (list (quote red) 700) (list (quote orange) 600) … (list (quote violet) 400)) red700 orange600 violet400

More Syntactic Sugar To the reader, ’pi is exactly the same as if you had typed (quote pi) Remember REPL 'pi ;Value: pi User types ’pi (quote pi) read pi eval pi print

More Syntactic Sugar To the reader, ’pi is exactly the same as if you had typed (quote pi) Remember REPL 'pi ;Value: pi '17 ;Value: 17 '"hi there" ;Value: "hi there" User types ’17 (quote 17) read 17 eval 17 print

More Syntactic Sugar To the reader, ’pi is exactly the same as if you had typed (quote pi) Remember REPL 'pi ;Value: pi '17 ;Value: 17 '"hi there" ;Value: "hi there" '(+ 3 4) ;Value: (+ 3 4) User types ’(+ 3 4) (quote (+ 3 4)) read (+ 3 4) eval (+ 3 4) print

More Syntactic Sugar To the reader, ’pi is exactly the same as if you had typed (quote pi) Remember REPL 'pi ;Value: pi '17 ;Value: 17 '"hi there" ;Value: "hi there" '(+ 3 4) ;Value: (+ 3 4) ''pi ;Value: (quote pi) But in Dr. Scheme, 'pi User types ’’pi (quote (quote pi)) read (quote pi) eval (quote pi) print

But wait… Clues about “guts” of Scheme (pair? (quote (+ 2 3))) ;Value: #t (pair? '(+ 2 3)) ;Value: #t (car '(+ 2 3)) ;Value: + (cadr '(+ 2 3)) ;Value: 2 (null? (cdddr '(+ 2 3))) ;Value: #t +23 Now we know that expressions are represented by lists!

Your turn: what does evaluating these print out? (define x 20) (+ x 3) ==> '(+ x 3) ==> (list (quote +) x '3) ==> (list '+ x 3) ==> (list + x 3) ==>

Your turn: what does evaluating these print out? (define x 20) (+ x 3) ==> '(+ x 3) ==> (list (quote +) x '3) ==> (list '+ x 3) ==> (list + x 3) ==> 23 (+ x 3) (+ 20 3) ([procedure #…] 20 3)

Grimson’s Rule of Thumb for Quote '(quote fred (quote quote) (+ 3 5)) (quote (quote fred (quote quote) (+ 3 5))) ???

The Grimson Rule of Thumb for Quote '((quote fred) (quote quote) (+ 3 5))) (quote ((quote fred) (quote quote) (+ 3 5)))) ??? What's the value of the quoted expression? WHATEVER IS UNDER YOUR THUMB! ('fred 'quote (+ 3 5))) '

Revisit making list structure with symbols (list (list (quote red) 700) (list (quote orange) 600) … (list (quote violet) 400)) (list (list ’red 700) (list ’orange 600) … (list ’violet 400)) ’((red 700) (orange 600) (yellow 575) (green 550) (cyan 510) (blue 470) (violet 400)) Because the reader knows how to turn parenthesized (for lists) and dotted (for pairs) expressions into list structure! red700 orange600 violet400

Aside: What all does the reader “know”? Recognizes and creates Various kinds of numbers –312 ==> integer –3.12e17 ==> real, etc. Strings enclosed by “…” Booleans #t and #f Symbols ’… ==> (quote …) (…) ==> pairs (and lists, which are made of pairs) and a few other obscure things

Traditional LISP structure: association list A list where each element is a list of the key and value. 15 x 20 y x: 15 y: 20 Represent the table as the alist: ((x 15) (y 20))

Alist operation: find-assoc (define (find-assoc key alist) (cond ((null? alist) #f) ((equal? key (caar alist)) (cadar alist)) (else (find-assoc key (cdr alist))))) (define a1 '((x 15) (y 20))) (find-assoc 'y a1) ==> x 20 y

An aside on testing equality = tests equality of numbers Eq?Tests equality of symbols Equal?Tests equality of symbols, numbers or lists of symbols and/or numbers that print the same

Alist operation: add-assoc (define (add-assoc key val alist) (cons (list key val) alist)) (define a2 (add-assoc 'y 10 a1)) a2 ==> ((y 10) (x 15) (y 20)) (find-assoc 'y a2) ==> 10 We say that the new binding for y “shadows” the previous one

Alists are not an abstract data type Missing a constructor: Used quote or list to construct (define a1 '((x 15) (y 20))) There is no abstraction barrier: the implementation is exposed. User may operate on alists using standard list operations. (filter (lambda (a) ( ((x 15))

Why do we care that Alists are not an ADT? Modularity is essential for software engineering Build a program by sticking modules together Can change one module without affecting the rest Alists have poor modularity Programs may use list ops like filter and map on alists These ops will fail if the implementation of alists change Must change whole program if you want a different table To achieve modularity, hide information Hide the fact that the table is implemented as a list Do not allow rest of program to use list operations ADT techniques exist in order to do this

Symbolic differentiation (deriv ) ==> Algebraic expressionRepresentation x + 3(+ x 3) xx 5y(* 5 y) x + y + 3(+ x (+ y 3)) (deriv '(+ x 3) 'x) ==> 1 (deriv '(+ (* x y) 4) 'x) ==> y (deriv '(* x x) 'x) ==> (+ x x)

Building a system for differentiation Example of: Lists of lists How to use the symbol type Symbolic manipulation 1. how to get started 2. a direct implementation 3. a better implementation

1. How to get started Analyze the problem precisely deriv constant dx = 0 deriv variable dx = 1 if variable is the same as x = 0 otherwise deriv (e1+e2) dx = deriv e1 dx + deriv e2 dx deriv (e1*e2) dx = e1 * (deriv e2 dx) + e2 * (deriv e1 dx) Observe: e1 and e2 might be complex subexpressions derivative of (e1+e2) formed from deriv e1 and deriv e2 a tree problem

Type of the data will guide implementation legal expressions x(+ x y) 2(* 2 x)(+ (* x y) 3) illegal expressions *(3 5 +)(+ x y z) ()(3)(* x) ; Expr = SimpleExpr | CompoundExpr ; SimpleExpr = number | symbol ; CompoundExpr = a list of three elements where the first element is either + or * ; = pair >>

2. A direct implementation Overall plan: one branch for each subpart of the type (define deriv (lambda (expr var) (if (simple-expr? expr) ))) To implement simple-expr? look at the type CompoundExpr is a pair nothing inside SimpleExpr is a pair therefore (define simple-expr? (lambda (e) (not (pair? e))))

Simple expressions One branch for each subpart of the type (define deriv (lambda (expr var) (if (simple-expr? expr) (if (number? expr) ) ))) Implement each branch by looking at the math 0 (if (eq? expr var) 1 0)

Compound expressions One branch for each subpart of the type (define deriv (lambda (expr var) (if (simple-expr? expr) (if (number? expr) 0 (if (eq? expr var) 1 0)) (if (eq? (car expr) '+) ) )))

Sum expressions To implement the sum branch, look at the math (define deriv (lambda (expr var) (if (simple-expr? expr) (if (number? expr) 0 (if (eq? expr var) 1 0)) (if (eq? (car expr) '+) (list '+ (deriv (cadr expr) var) (deriv (caddr expr) var)) ) ))) (deriv '(+ x y) 'x) ==> (+ 1 0) (a list!)

The direct implementation works, but... Programs always change after initial design Hard to read Hard to extend safely to new operators or simple exprs Can't change representation of expressions Source of the problems: nested if expressions explicit access to and construction of lists few useful names within the function to guide reader

3. A better implementation 1. Use cond instead of nested if expressions 2. Use data abstraction do this for every branch : (define variable? (lambda (e) (and (not (pair? e)) (symbol? e)))) To use cond: write a predicate that collects all tests to get to a branch: (define sum-expr? (lambda (e) (and (pair? e) (eq? (car e) '+)))) ; type: Expr -> boolean

Use data abstractions To eliminate dependence on the representation: (define make-sum (lambda (e1 e2) (list '+ e1 e2)) (define addend (lambda (sum) (cadr sum))) (define augend (lambda (sum) (caddr sum)))

A better implementation (define deriv (lambda (expr var) (cond ((number? expr) 0) ((variable? expr) (if (eq? expr var) 1 0)) ((sum-expr? expr) (make-sum (deriv (addend expr) var) (deriv (augend expr) var))) ((product-expr? expr) ) (else (error "unknown expression type" expr)) ))

Isolating changes to improve performance (deriv '(+ x y) 'x) ==> (+ 1 0) (a list!) (define make-sum (lambda (e1 e2) (cond ((number? e1) (if (number? e2) (+ e1 e2) (list '+ e1 e2))) ((number? e2) (list '+ e2 e1)) (else (list '+ e1 e2))))) (deriv '(+ x y) 'x) ==> 1

Modularity makes changes easier But conventional mathematics doesn’t use prefix notation like this: (+ 2 x) or (* (+ 3 x) (+ x y)) Could we change our program somehow to use more algebraic expressions, still fully parenthesized, like: (2 + x) or ((3 + x) * (x + y)) What do we need to change?

Just change data abstraction Constructors Accessors Predicates (define (make-sum e1 e2) (list e1 '+ e2)) (define (augend expr) (caddr expr)) (define (sum-expr? expr) (and (pair? expr) (eq? '+ (cadr expr))))

Separating simplification from differentiation Exploit Modularity: Rather than changing the code to handle simplification of expressions, write a separate simplifier (define (simplify expr) (cond ((or (number? expr) (variable? expr)) expr) ((sum-expr? expr) (simplify-sum (simplify (addend expr)) (simplify (augend expr)))) ((product-expr? expr) (simplify-product (simplify (multiplier expr)) (simplify (multiplicand expr)))) (else (error "unknown expr type" expr))))

Simplifying sums (define (simplify-sum add aug) (cond ((and (number? add) (number? aug)) ;; both terms are numbers: add them (+ add aug)) ((or (number? add) (number? aug)) ;; one term only is number (cond ((and (number? add) (zero? add)) aug) ((and (number? aug) (zero? aug)) add) (else (make-sum add aug)))) ((eq? add aug) ;; adding same term twice (make-product 2 add)) … (+ 2 3)  5 (+ 2 x)  (+ 2 x) (+ 0 x)  x (+ x 0)  x (+ x x)  (* 2 x)

More special cases in simplification (define (simplify-sum add aug) (cond … ((product-expr? aug) ;; check for special case of (+ x (* 3 x)) ;; i.e., adding something to a multiple of itself (let ((mulr (simplify (multiplier aug))) (muld (simplify (multiplicand aug)))) (if (and (number? mulr) (eq? add muld)) (make-product (+ 1 mulr) add) ;; not special case: lose (make-sum add aug)))) (else (make-sum add aug)))) (+ x (* 3 x))  (* 4 x)

Special cases in simplifying products (define (simplify-product f1 f2) (cond ((and (number? f1) (number? f2)) (* f1 f2)) ((number? f1) (cond ((zero? f1) 0) ((= f1 1) f2) (else (make-product f1 f2)))) ((number? f2) (cond ((zero? f2) 0) ((= f2 1) f1) (else (make-product f2 f1)))) (else (make-product f1 f2)))) (* (+ 3 x) 2)  (* 2 (+ 3 x)) (* 3 5)  15 (* 0 (+ x 1))  0 (* 1 (+ x 1))  (+ x 1)

Simplified derivative looks better (deriv '(+ x 3) 'x) ;Value: (+ 1 0) (deriv '(+ x (* x y)) 'x) ;Value: (+ 1 (+ (* x 0) (* 1 y))) (simplify (deriv '(+ x 3) 'x)) ;Value: 1 (simplify (deriv '(+ x (* x y)) 'x)) ;Value: (+ 1 y) But, which is simpler? a*(b+c) or a*b + a*c Depends on context…

Recap Symbols Are first class objects Allow us to represent names Quotation (and the reader’s syntactic sugar for ') Let us evaluate (quote …) to get … as the value –I.e., “prevents one evaluation” –Not really, but informally, has that effect. Lisp expressions are represented as lists Encourages writing programs that manipulate programs –Much more, later Symbolic differentiation (introduction)